All categories

3 years ago

Need help ASAP!! I'll give brainliest to correct answer!!!

Mathematics

2 answers:

erastovalidia [21]3 years ago

4 0

Step-by-step explanation:

so...the answer is C.27.7m

Diano4ka-milaya [45]3 years ago

4 0

Answer:

27.7

................

You might be interested in

15.09 ÷ 0.01 = ? 267.4 ÷ 1/10 = ?

Masja [62]

Answer:

Dividing by a fraction is multiplying by the reciprocal

15.09/1/100=15.09*100=1509

267.4/1/10=267.4*10=2674

8 0

3 years ago

Read 2 more answers

Secant DB intersects secant DZ at point D. Find the length of DY

netineya [11]

By the Intersecting Secants Theorem we have
2x(2x+7)=x(x+23)
4x^2+14x=x^2+23x
3x^2=9x
3x=9
x=3
Since DY is length 2x,
the length of DY is 6.

5 0

3 years ago

What is the value of the expression when a = -1.4 and b = -2.7?

Nana76 [90]

Answer:

The correct answer is A) 1.3

Step-by-step explanation:

To find this, simply input the correct values.

-|a| - b

-|-1.4| - -2.7

-1.4 + 2.7

1.3

3 0

3 years ago

Read 2 more answers

Help pls urgent!!!!!!!!!!

Nesterboy [21]

Answer:

Step-by-step explanation:

8 0

4 years ago

A rat is trapped in a maze. Initially he has to choose one of two directions. If he goes to the right, then he will wander aroun

Brut [27]

Answer:

The expected number of minutes the rat will be trapped in the maze is 21 minutes.

Step-by-step explanation:

The rat has two directions to leave the maze.

The probability of selecting any of the two directions is, $\frac{1}{2}$ .

If the rat selects the right direction, the rat will return to the starting point after 3 minutes.

If the rat selects the left direction then the rat will leave the maze with probability $\frac{1}{3}$ after 2 minutes. And with probability $\frac{2}{3}$ the rat will return to the starting point after 5 minutes of wandering.

Let X = number of minutes the rat will be trapped in the maze.

Compute the expected value of X as follows:

$E(X)=[(3+E(X)\times\frac{1}{2} ]+[2\times\frac{1}{6} ]+[(5+E(X)\times\frac{2}{6} ]\\E(X)=\frac{3}{2} +\frac{E(X)}{2}+\frac{1}{3}+\frac{5}{3} +\frac{E(X)}{3} \\E(X)-\frac{E(X)}{2}-\frac{E(X)}{3}=\frac{3}{2} +\frac{1}{3}+\frac{5}{3} \\\frac{6E(X)-3E(X)-2E(X)}{6}=\frac{9+2+10}{6}\\\frac{E(X)}{6}=\frac{21}{6}\\E(X)=21$

Thus, the expected number of minutes the rat will be trapped in the maze is 21 minutes.

3 0

4 years ago